Answer:
Since RANDY operates randomly, any file within the specified index range will have the recurrence relation as follows:
T(n) = T(n-i) + O(1)
Here, the probability is 1/n, where i can vary between 1 and n. The variable n in T(n) denotes the size of the index range, which will subsequently reduce to (n-i) in the following iteration.
Given that i is probabilistically distributed from 1 to n, the average case time complexity can then be expressed as:
T(n) = 
Next, solving T(n) = T(n/2) + O(1)
yields T(n) = O(log n).
Thus, the complexity of this algorithm is O(log n).
It should be noted that this represents the average time complexity due to the algorithm's randomized nature. In the worst-case scenario, the index range may only decrease by 1, resulting in a time complexity of O(n) since the worst-case scenario would be T(n) = T(n-1) + O(1).
Answer:
cache I guess
Explanation:
unsure or memory HDD or SSD
The decimal number 8620 can be expressed in the following ways:
a = 1000
0110 0010 0000
b = 1011
1001 0101 0011
c = 1110
1100 0010 0000
d =
10000110101100
I hope these representations answer your question and prove useful.
Response: explained in the explanation section
Explanation:
Given that:
Assume D(k) =║ true if [1::: k] is a valid sequence of words, or false otherwise
the sub problem s[1::: k] is a valid sequence of words IFF s[1::: 1] is valid and s[ 1 + 1::: k] is a valid word.
Thus, we derive that D(k) is defined by the following recurrence relation:
D(k) = ║ false max(d[l] ∧ DICT(s[1 + 1::: k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1::: k] ∦ array of boolean variables.
for a ← 1 to m
do;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1::: a])
d (a) ← True
(b). Algorithm Output
if D[k] == True
stack = temp stack ∦stack assists in displaying the strings in order
c = k
while C > 0
stack push (s [w(c)]::: C] // w(p) denotes the index in s[1::: k] of the valid word // at position c
P = W (p) - 1
output stack
= 0 =
cheers, I hope this aids you!!!
The data for the middle node shows as –250.... Please outline the pseudocode to substitute the middle node in the linked list. Assume that the head pointer is designated as Head_ptr and the entry data for the new node is defined as Entry.
